= Solution
\b[The ultrafilter space.] An <ultrafilter> $U$ on $\mathbb N$ is a proper <filter on a set>: it is upward closed, closed under finite intersections, excludes $\varnothing$, and contains exactly one of $A,A^c$ for each $A\subseteq\mathbb N$. The <Stone-Čech compactification of the natural numbers> is the set of these <ultrafilters>, with basic sets
$$
\widehat A=\{U:A\in U\}.
$$
They form a basis since $\widehat A\cap\widehat B=\widehat{A\cap B}$ and $\widehat{\mathbb N}=\beta\mathbb N$. Moreover $(\widehat A)^c=\widehat{A^c}$, so each is a <clopen set>. The <principal ultrafilter> at $n$ embeds $\mathbb N$ in this space; every nonempty $\widehat A$ contains such a point, so the embedded copy is <dense>.
Distinct <ultrafilters> $U,V$ have some $A\in U$ with $A\notin V$. Then $A^c\in V$, and $\widehat A,\widehat{A^c}$ are disjoint <open sets> separating them. Thus the space is <Hausdorff>.
For <compactness>, any <open set> cover can be refined to a cover by basic sets. If $\{\widehat{A_i}\}$ has no finite subcover, the sets $A_i^c$ have the <finite intersection property>: otherwise finitely many $A_i$ cover $\mathbb N$, and their basic sets cover every <ultrafilter>. Extend the proper <filter on a set> generated by these complements to an <ultrafilter> $U$. This extension follows from <Zorn lemma>: unions of chains of proper filters remain proper, and a maximal proper filter decides every $A$ versus $A^c$. Indeed, if adding $A$ made the filter improper, some existing filter member would be disjoint from $A$, forcing $A^c$ into the filter. This $U$ contains every $A_i^c$ and misses every $\widehat{A_i}$, a contradiction. Hence
$$
\boxed{\beta\mathbb N\text{ is compact and Hausdorff}.}
$$
This proves the <compact Hausdorff topology on ultrafilters> directly.
\b[An idempotent.] The <addition on the Stone-Čech compactification of the natural numbers> can be defined by
$$
A\in U+V\quad\Longleftrightarrow\quad\{n:A-n\in V\}\in U,\qquad A-n=\{k:n+k\in A\}.
$$
Associativity is supplied. For fixed $V$, the inverse image of $\widehat A$ under $U\mapsto U+V$ is $\widehat{\{n:A-n\in V\}}$, so this translation is <continuous> directly from the definition. This makes a <left-topological semigroup> in the convention used here.
We prove the <Ellis–Numakura lemma> in this case. Among nonempty compact <subsemigroups> choose one, $K$, minimal by inclusion. It exists by <Zorn lemma>: a descending chain has nonempty intersection by <compactness>, and that intersection is again a compact <subsemigroup>. For $a\in K$, the set $K+a$ is a nonempty compact <subsemigroup> contained in $K$. <Compactness> follows from <continuity>; closure under addition follows from
$$
(x+a)+(y+a)=(x+a+y)+a\in K+a.
$$
Minimality gives $K+a=K$. Hence there is $b\in K$ with $b+a=a$. The set
$$
L=\{x\in K:x+a=a\}
$$
is nonempty, compact, and a <subsemigroup>: if $x,y\in L$, then $(x+y)+a=x+(y+a)=x+a=a$. Thus minimality gives $L=K$, and in particular $a+a=a$. Therefore
$$
\boxed{\text{there is an idempotent ultrafilter in }\beta\mathbb N.}
$$
This is an <idempotent ultrafilter on the natural numbers>.
\b[Isolating the sequence terms.] Now use the hypothetical <convergent sequence> of distinct <ultrafilters>. For each $n$, choose $C_n\in U_n$ with $C_n^c\in U$, possible since $U_n\ne U$. Convergence implies $C_n^c\in U_m$ for all sufficiently large $m$. Thus only finitely many indices $m\ne n$ have $C_n\in U_m$. Call this finite set $E_n$.
For each $m\in E_n$, choose $D_{n,m}\in U_n$ with $D_{n,m}\notin U_m$, using distinctness. Set
$$
A_n=C_n\cap\bigcap_{m\in E_n}D_{n,m}.
$$
Finite-intersection closure puts $A_n$ in $U_n$. For $m\ne n$, either $C_n\notin U_m$ or one of the chosen $D_{n,m}$ is absent from $U_m$; upward closure then prevents $A_n$ from belonging to $U_m$. Hence
$$
\boxed{A_n\in U_m\quad\Longleftrightarrow\quad m=n.}
$$
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